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The probability distribution of a random variable X is given below : ( array |c|c|c|c|c|c|c|c|c| X & 4k & 30 7 k & 32 7 k & 34 7 k & 36 7 k & 38 7 k & 40 7 k & 6k P(X) & 2 15 & 1 15 & 2 15 & 1 5 & 1 15 & 2 15 & 1 5 & 1 15 array ) If E ( X )= 263 15 , then P ( X <20) is equal to :

Options

  1. A11 15
  2. B3 5
  3. C14 15
  4. D8 15

Correct answer

A. 11 15

Step-by-step solution

E(X) = X_i P(X_i) = 526k 15 7 = 263 15 k = 7 2 Substituting k = 7 2 , the values of X become: X : 14, 15, 16, 17, 18, 19, 20, 21 P(X) : 2 15 , 1 15 , 2 15 , 1 5 , 1 15 , 2 15 , 1 5 , 1 15 P(X = 2 15 + 1 15 + 2 15 + 1 5 + 1 15 + 2 15 = 11 15

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